# Solve the double angle equation

• Feb 27th 2011, 03:31 AM
Fionafaye
Solve the double angle equation
Im stuck on this question in my maths homework:

Solve the equation √3tan2x-1=0 for 0<X<π

Here is what i've done:

√3tan2x-1=0
tan2x=1/√3
Using trig identities,
tanx= sinx/cosx
so tan2x=sin2x/cos2x < is this right?
I've then used the trig expansions:
2sinxcosx/1-2sin^2x < Am I using the right expansion for cos2x?

Am I using the right method? I have tried to go further than this, but nothing has worked out, so where do I go from here?

• Feb 27th 2011, 03:48 AM
e^(i*pi)
Don't bother changing it to sin and cos - arctan is a fine operation and you can use it on your second line

$\arctan(\tan(2x)) = \arctan\left({\dfrac{1}{\sqrt{3}}\right) \implies 2x = \dfrac{\pi}{6} \implies x = \dfrac{\pi}{12}$

That is the principle solution but remember that since you have a double angle the period is $\dfrac{\pi}{2}$ so you'll need to add that to get the second solution
• Feb 27th 2011, 03:52 AM
Fionafaye
I would do that, but we havent been taught it in class and its not in our notes, so I have to do it the sin and cos way. Thanks anyway. :)
• Feb 27th 2011, 04:12 AM
Plato
Quote:

Originally Posted by Fionafaye
I would do that, but we havent been taught it in class and its not in our notes, so I have to do it the sin and cos way.

Draw an equilateral triangle with sides of length two.
Draw an altitude from the summit angle.
It is also a median and angular bisector.
The altitude has length $\sqrt3$ and the adjunct side has length $1$.
So half the summit angle is $\frac{\pi}{6}$ and $\tan\left(\frac{\pi}{6}\right)= \left(\frac{1}{\sqrt3}\right)$.
Now finish.
• Feb 27th 2011, 04:35 AM
sa-ri-ga-ma
If you want it in sin and cos, then

√3sin(2x) = cos(2x)

sin^2(x) + 2√3sin(x)cos(x) - cos^2(x) = 0

Solve the quadratic for sin(x). From that find tan(x)
• Feb 28th 2011, 03:23 AM
teddystanley
a tricky kind of question.